Submitting JSON data via JQuery ajax.post to PHP

Question

Im submitting Data to a php file via AJAX using POST. It worked fine with just submitting strings, but now I wanted to submit my JS Object with JSON and decode it on PHP side.

In the console I can see, that my data is submitted correctly but on PHP side json_decode returns NULL.

I've tried the following:

this.getAbsence = function()
{
    alert(JSON.stringify(this));
    jQuery.ajax({
        type: "POST",
        contentType: "application/json; charset=utf-8",
        url: "ajax/selectSingle.php?m=getAbsence",
        data: JSON.stringify(this),
        success : function(data){
            alert(data);
        }
    });
}

PHP:

echo $_POST['data'];
echo json_decode($_POST['data']);
echo var_dump(json_decode($_POST['data']));

And:

this.getAbsence = function()
{
    alert(JSON.stringify(this));
    jQuery.ajax({
        type: "POST",
        contentType: "application/json; charset=utf-8",
        url: "ajax/selectSingle.php?m=getAbsence",
        data: {'Absence' : JSON.stringify(this)},
        success : function(data){
            alert(data);
        }
    });
}

PHP:

echo $_POST['Absence'];
echo json_decode($_POST['Absence']);
echo var_dump(json_decode($_POST['Absence']));

The alert was just to check everything is alright...

And yea usual string were echoed correctly :-)

Im sorry about that, but of course I have to try these answers, and all of them were not right... But yeah I found my answer - but in my opinion I do help if I am commenting why the answer did not help me; if I am not commenting at all the people do not know, if I tried them and if the right answer is already given. — SamiSalami
– SamiSalami, Commented Jun 8, 2012 at 12:12

VeeeneX · Accepted Answer · 2014-05-06 15:41:04Z

Where you went wrong in your code in the first code is that you must have used this:

var_dump(json_decode(file_get_contents("php://input"))); //and not $_POST['data']

Quoting from PHP Manual

php://input is a read-only stream that allows you to read raw data from the request body.

Since in your case, you are submitting a JSON in the body, you have to read it from this stream. Usual method of $_POST['field_name'] wont work, because the post body is not in an URLencoded format.

In the second part, you must have used this:

contentType: "application/json; charset=utf-8",
url: "ajax/selectSingle.php?m=getAbsence",
data: JSON.stringify({'Absence' : JSON.stringify(this)}),

UPDATE:

When request has a content type application/json, PHP wont parse the request and give you the JSON object in $_POST, you must parse it yourself from the raw HTTP body. The JSON string is retrieved using file_get_contents("php://input");.

If you must get that using $_POSTyou would make it:

data: {"data":JSON.stringify({'Absence' : JSON.stringify(this)})},

And then in PHP do:

$json = json_decode($_POST['data']);

Thank you very much, that did it! Tried your answer at last ^^ Though I am a bit confused because I read examples where JSON objects from an ajax call where just echoed from a $_POST... But really thanks :)
Wow, thank you very much with that update it is much more understandable :)
file_get_contents("php://input"); is the one i was looking for. Spent hours before I find this

YemSalat · Accepted Answer · 2012-06-08 10:58:02Z

0

Single quotes are not valid for php's json_encode, use the double quotes for both field names and values.

answered Jun 8, 2012 at 10:58

YemSalat

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5 Comments

Flygenring Over a year ago

I don't see how this is relevant at all, as all data is transferred as part of the POST-request and the JSON is formatted by the JSON.stringify-function - single quotes are only used in the object with settings used by the ajax-function.

YemSalat Over a year ago

data: {'Absence' : JSON.stringify(this)} - here, php receives the whole thing, including 'Absence', which should also be in double quotes, otherwise it is not valid.

SamiSalami Over a year ago

No, "Absence" is not a property of the given json-object, I am just adressing it with "Absence" to get it via $_POST['Absence'], it ist not a property so it does not have to be in double quotes

YemSalat Over a year ago

As I said - field names should also be in double quotes. See example #3: php.net/manual/en/function.json-decode.php

SamiSalami Over a year ago

Yeah, but it is not a fieldname, there will be something submitted like: Absence = aid:1, name:bla... (just pseudo-code) and in fulltext: data: {'Absence':{"aid":1, "name": bla}}; Absence is not part of the object it is the given name

Flygenring · Accepted Answer · 2012-06-08 11:41:14Z

0

To me, it looks like you should reformat your AJAX object. The url-property should only be the URL for the target php-file and any data that needs to be posted should be in the form of a query-string in the data-property. The following should work as you expected:

this.getAbsence = function() {
  var strJSONData = JSON.stringify(this);
  alert(strJSONData);
  jQuery.ajax({
    type: 'POST',
    contentType: 'application/json; charset=utf-8',
    url: 'ajax/selectSingle.php',
    data: 'm=getAbsence&Absence=' + strJSONData,
    success: function(data) {
      alert(data);
    }
  });
}

edited Jun 8, 2012 at 11:41

answered Jun 8, 2012 at 11:11

Flygenring

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4 Comments

SamiSalami Over a year ago

strJSONData looks fine, but I am just getting empty strings with "echo $_GET['Absence'];", "echo json_decode($_GET['Absence'])" - also tried with post, but this is a submission as a GET... I thought of doing it via POST or am I misunderstanding something related to the submission of JSON-objects via Ajax? Do they have to be submitted as a GET (url param)?

Flygenring Over a year ago

It should give you empty strings when you access the $_GET variable on a POST request. You could either use the $_POST or the $_REQUEST variable to access the submitted data (etc).

SamiSalami Over a year ago

As I told you o tried POST as well, but i tried GET first, because, you are submitting the data like url params, but as I said maybe I am just misunderstanding and the ajay post is kinda a fake "POST" :-)

Flygenring Over a year ago

Ohh I'm sorry, I missed that part of your comment. It's the type that determines how to retrieve the data in your php-script, no matter how the data is formatted – it's just converted to a query-string anyways according to the docs.

Rishabh · Accepted Answer · 2012-06-08 11:48:40Z

0

try this

  var vThis = this;
  this.getAbsence = function()
  {
    alert(JSON.stringify(vThis));
    jQuery.ajax({
       type: "POST",
       contentType: "application/json; charset=utf-8",
       url: "ajax/selectSingle.php?m=getAbsence",
       data: JSON.stringify(vThis),
       success : function(data){
         alert(data);
       } 
     });
   }

EDIT

I think we can also do this!

  var vThis = this;
  this.getAbsence = function()
  {
    alert(JSON.stringify(vThis));
    jQuery.ajax({
       type: "POST",
       dataType: "json",
       url: "ajax/selectSingle.php?m=getAbsence",
       data: vThis,
       success : function(data){
         alert(data);
       } 
     });
   }

and in PHP

print_r($_POST);

edited Jun 8, 2012 at 11:48

answered Jun 8, 2012 at 11:16

Rishabh

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5 Comments

Flygenring Over a year ago

The reference this in the function body should be a reference to the object on which the function was called, so that part should work even if it's not exactly pretty or readable code

Rishabh Over a year ago

yes, alert will work, but the this in jQuery.ajax will not be the reference to the object on which the function was called

SamiSalami Over a year ago

One good point u have, that this might not be the object anymore, though I do not think so, because I have seen other examples where this was used directly. Your second suggestions leads to an erro in the console: "too much recursion" - Don't know why, my browser even freezes for a moment x) the first one by print_r($_POST) I am only getting "Array()" of course and $_POST['data'] ist empty :-/

Rishabh Over a year ago

are you binding this.getAbsence to the DOM ? If so, then second option will not work.

SamiSalami Over a year ago

No, I am not :-/ function Absence(){this.aid;this.eid;...[my method]...} in another js: myAbsence = new Absence(); myAbsence.aid = 1; myAbsence.getAbsence(); - But anyways I found my answer, but maybe it helps u or u have something so say sumething like "OMG, what are u doin?" ^^

Cristian Oana · Accepted Answer · 2022-02-23 14:40:59Z

0

On PHP side try this:

$sectionValue = htmlspecialchars($_POST['sectionValue'], ENT_QUOTES);
$dataToWrite = json_decode(html_entity_decode($sectionValue, ENT_QUOTES, "utf-8" ), true);

answered Feb 23, 2022 at 14:40

Cristian Oana

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